What is the solution for y^{''}-y^'+1/4 y=5+e^{x/2} ?

The solution for y^{''}-y^'+1/4 y=5+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+20+(e^{x/2}x^2)/2

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y^{''}-y^'+1/4 y=5+e^{x/2}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 5 + e^{\frac{x}{2}}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 20 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

Solution steps

y^{''} (x) - y^{'} (x) + \frac{1}{4} y = 5 + e^{\frac{x}{2}}

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 20 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{\frac{x}{2}} x + 20 + \frac{e ^{\frac{x}{2}} x ^{2}}{2}

x^{2} (1 - 8 x)

\frac{d N}{d t} + N = Nt e^{t + 8}

\frac{d}{d x} (\frac{1 + x ^{3}}{1 - x ^{3}})

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What is the solution for y^{''}-y^'+1/4 y=5+e^{x/2} ?
The solution for y^{''}-y^'+1/4 y=5+e^{x/2} is y=c_{1}e^{x/2}+c_{2}e^{x/2}x+20+(e^{x/2}x^2)/2